1 `cm^(3)` of waterr at its boiling point absorbs 540 calories of heat to become steam with a volume of 1671 `cm^(3)`. If the atmospheric pressure ` = 1.013 xx 10^(5) N//m^(2)` and the mechanical equivalent of heat ` = 4.19` J/calorie, the energy spent in this process in overcoming inter molecular forces is

Options

A. 540 calorie
B. 40 calorie
C. 500 calorie
D. zero

Explanation

To solve the problem, we need to find the energy spent in overcoming intermolecular forces during the phase change of water to steam. This can be calculated using the first law of thermodynamics, which relates the heat absorbed, work done, and change in internal energy.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Heat absorbed (Q) = 540 calories
- Initial volume of water (V_initial) = 1 cm³
- Final volume of steam (V_final) = 1671 cm³
- Atmospheric pressure (P) = 1.013 × 10^5 N/m²
- Mechanical equivalent of heat = 4.19 J/calorie

2. **Calculate the Change in Volume (ΔV):**
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 1671 \, \text{cm}^3 - 1 \, \text{cm}^3 = 1670 \, \text{cm}^3 \]
Convert ΔV to m³:
\[ \Delta V = 1670 \, \text{cm}^3 \times 10^{-6} \, \text{m}^3/\text{cm}^3 = 1.67 \times 10^{-3} \, \text{m}^3 \]

3. **Calculate the Work Done (W):**
The work done by the system during the expansion at constant pressure is given by:
\[ W = P \Delta V \]
Substituting the values:
\[ W = (1.013 \times 10^5 \, \text{N/m}^2) \times (1.67 \times 10^{-3} \, \text{m}^3) = 168.1711 \, \text{J} \]

4. **Convert Work Done from Joules to Calories:**
To convert Joules to calories, use the conversion factor:
\[ 1 \, \text{calorie} = 4.19 \, \text{J} \]
Therefore:
\[ W = \frac{168.1711 \, \text{J}}{4.19 \, \text{J/calorie}} \approx 40.1 \, \text{calories} \]

5. **Calculate the Change in Internal Energy (ΔU):**
According to the first law of thermodynamics:
\[ Q = \Delta U + W \]
Rearranging gives:
\[ \Delta U = Q - W \]
Substituting the values:
\[ \Delta U = 540 \, \text{calories} - 40.1 \, \text{calories} \approx 499.9 \, \text{calories} \]

6. **Final Result:**
The energy spent in overcoming intermolecular forces is approximately:
\[ \Delta U \approx 500 \, \text{calories} \]

### Conclusion:
The energy spent in overcoming intermolecular forces during the phase change of water to steam is approximately **500 calories**.